Infinite number of charges of `4mu C`, respectively are placed along the X-axis at 1 m, 2 m, 4 m, 8 m and So. The coulomb field at origin for all these charges is:

Options

A. `4.8 xx 10^(6)(N)/(C)`
B. `4.8 xx 10^(5)(N)/(C)`
C. `4.8 xx 10^(4)(N)/(C)`
D. `4.8 xx 10^(3)(N)/(C)`

Explanation

To find the electric field at the origin due to an infinite number of charges of \(4 \mu C\) placed along the X-axis at distances of \(1 m\), \(2 m\), \(4 m\), \(8 m\), and so on, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Charge and Distance**:
Each charge \(q = 4 \mu C = 4 \times 10^{-6} C\) is placed at distances \(r = 1 m, 2 m, 4 m, 8 m, \ldots\) from the origin.

2. **Electric Field due to a Point Charge**:
The electric field \(E\) due to a point charge is given by the formula:
\[ E = \frac{k \cdot q}{r^2} \]
where \(k\) is Coulomb's constant (\(k \approx 9 \times 10^9 \, N m^2/C^2\)).

3. **Calculate the Electric Field Contributions**:
The total electric field at the origin due to all the charges can be calculated by summing the contributions from each charge:
\[ E_{\text{total}} = \sum_{n=0}^{\infty} \frac{k \cdot q}{(2^n)^2} \]
Here, the distances are \(1, 2, 4, 8, \ldots\) which can be represented as \(2^n\) for \(n = 0, 1, 2, \ldots\).

4. **Substituting Values**:
Substitute \(q\) and \(k\) into the equation:
\[ E_{\text{total}} = k \cdot q \sum_{n=0}^{\infty} \frac{1}{(2^n)^2} = k \cdot q \sum_{n=0}^{\infty} \frac{1}{4^n} \]

5. **Sum of the Infinite Series**:
The series \(\sum_{n=0}^{\infty} \frac{1}{4^n}\) is a geometric series with the first term \(a = 1\) and common ratio \(r = \frac{1}{4}\):
\[ \text{Sum} = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

6. **Final Calculation**:
Now substitute back into the electric field equation:
\[ E_{\text{total}} = k \cdot q \cdot \frac{4}{3} = \left(9 \times 10^9 \, N m^2/C^2\right) \cdot \left(4 \times 10^{-6} C\right) \cdot \frac{4}{3} \]

7. **Calculate the Result**:
\[ E_{\text{total}} = 9 \times 10^9 \cdot 4 \times 10^{-6} \cdot \frac{4}{3} = 48 \times 10^3 \, N/C = 4.8 \times 10^4 \, N/C \]

### Conclusion
The electric field at the origin due to the infinite number of charges is:
\[ E = 4.8 \times 10^4 \, N/C \]